Problem: Factor the following expression: $3$ $x^2+$ $4$ $x$ $-15$
This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(-15)} &=& -45 \\ {a} + {b} &=& & & {4} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-45$ and add them together. Remember, since $-45$ is negative, one of the factors must be negative. The factors that add up to ${4}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-5}$ and ${b}$ is ${9}$ $ \begin{eqnarray} {ab} &=& ({-5})({9}) &=& -45 \\ {a} + {b} &=& {-5} + {9} &=& 4 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {3}x^2 {-5}x +{9}x {-15} $ Group the terms so that there is a common factor in each group: $ ({3}x^2 {-5}x) + ({9}x {-15}) $ Factor out the common factors: $ x(3x - 5) + 3(3x - 5) $ Notice how $(3x - 5)$ has become a common factor. Factor this out to find the answer. $(3x - 5)(x + 3)$